Question

if one zero of the polynomial is (a square + 9)x square + 13x +6a is raciprocal of tge other. Find the value of 'a'

Answer 1

Hi Mate !!


Let the Zeros be x and 1/x as given one zeros is reciprocal of other.


Given equation :- ( a² + 9 )x² + 13x + 6a

• Product of Zeros

$= \frac{constant \: term \: }{coeff \: of \: {x}^{2} }$

$x \times \frac{1}{x} = \frac{6a}{ {a}^{2} + 9}$


$1 = \frac{6a}{ {a}^{2} + 9 }$


a² + 9 = 6a

a² - 6a + 9 = 0

a² - 3a - 3a + 9 = 0

a ( a - 3 ) - 3 ( a - 3 ) = 0


( a - 3 ) ( a - 3 ) = 0

• ( a - 3 ) = 0

a = 3

• ( a - 3 ) = 0

a = 3

so, the value of a is 3

Answer 2

Step-by-step explanation:

problem solved...

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